How Do You Solve This Differential Equation with Initial Condition y(1)=3?

[tunabeast]

Homework Statement

$$x^2\frac{dy}{dx}=\frac{2sqrt(y)}{x^4}$$

Where y(1)=3

Homework Equations

The Attempt at a Solution

$$\int \frac{dy}{2sqrt(y)}=\int \frac{dx}{x^4}$$

$$4sqrt(y)=\frac{-1}{3x^3}+c$$

$$c=4sqrt(3)+\frac{1}{3}$$

So i end up with

$$4sqrt(y)=\frac{-1}{3x^3}+4sqrt(3)+\frac{1}{3}$$

I'v gone wrong somewhere as the solution given does not match, i just can't spot where my mistake is. Thanks in advance for any help

 

[dotman]

Hello,

How'd you get $$\int \frac{dy}{2\sqrt{y}}=\int \frac{dx}{x^4}$$
from

$$x^2\frac{dy}{dx}=\frac{2\sqrt{y}}{x^4}$$?

What happened to the $x^2$ term on the left?

 

[tunabeast]

Sorry about that, stupid error. In the initial problem it should be $$\frac{2sqrt(y)}{x^2}$$

 

[Dick]

Differentiate 4*sqrt(y). Do you get 1/(2*sqrt(y))?